Can you find the ordered pairs?

It is given that:

$\begin{cases} 2^{x^2+y} + 2^{x+y^2} = 128 & ...(1) \\ \sqrt{x} + \sqrt{y} = 2\sqrt{2} &...(2) \end{cases}$

From $(2)$ we note that $x \ge 0$ and $y \ge 0$ .

Using AM-GM inequality:

$\begin{aligned} 2^{x^2+y} + 2^{x+y^2} & \ge 2 \sqrt{2^{x^2+y + x+y^2}} \\ 128 & \ge 2 \sqrt{2^{x^2+y + x+y^2}} \\ \Rightarrow \sqrt{2^{x^2+y + x+y^2}} & \le 64 = 2^6 \\ 2^{x^2+x +y^2+y} & \le 2^{12} \\ x^2+x +y^2+y & \le 12 \end{aligned}$

Using Cauchy-Schwarz inequality:

$\begin{aligned} \left(\sqrt{x} + \sqrt{y} \right)^2 & \le (1+1)(x+y) \\ 8 & \le 2(x+y) \\ \Rightarrow x + y & \ge 4 \\ \Rightarrow (x+y)^2 & \le (1+1)(x^2+y^2) \\ x^2 + y^2 & \ge 8 \\ x^2+x +y^2+y & \ge 12\end{aligned}$

We note that $12 \le x^2+x +y^2+y \le 12 \quad \Rightarrow x^2+x +y^2+y = 12$ . And equality happens when $x = y$ and since $x + y = 4$ $\quad \Rightarrow x = y = 2$ . The sum of all real numbers $x,y$ satisfying both the equations is $2+2 = \boxed{4}$ .

I think it should be of number theory

$\sqrt{8}$ can be written as $2\sqrt{2}$ .

We see from the second equation that $x,y \ge 0$

Therefore $x=y=2$ or $x=8, y=0$ or $x=0, y=8$

We can apply these values in the first equation and see that only $x=y=2$ satisfy therefore our answer should be $2+2=\boxed{4}$